
This question asks for absolute value which can be found either in the Math menu or using catalog (2nd, zero) and abs comes up right away because it begins with an A :)




This question asks for absolute value which can be found either in the Math menu or using catalog (2nd, zero) and abs comes up right away because it begins with an A :)




Domain means x values and Range means y.
D comes before R in the alphabet and x is before y.
This problem gives us x values so use them to find y.
f(x) = 2x^2 – 8
f(-2) = 2(-2)^2 – 8 = 2(4) – 8 = 0 (This narrows it down to answers (3) or (4)!)
f(3) = 2(3)^2 – 8 = 2(9) – 8 = 10 (This makes the answer jump out at ya!)
f(5) = 2(5)^2 – 8 = 2(25) – 8 = 42
Can also use the table from the TI-84 :)


How did 6 – 2(x^2 – 1) become 6 – 2x^2 +2?
Help Evan identify this first step of his problem solving :)
That’s it! 2 points on the Regents!

2 points :)
Q: Leaving at 8:00 AM, if Ms. Brown drives at an average speed of 40 miles per hour, she will be late by 3 minutes. If she drives at an average speed of 60 miles per hour, she will be early by 3 minutes. How many miles per hour does Ms. Brown need to drive to get to work exactly on time?
A: 48 miles per hour
There is a 6 minute difference between the fast (60 mph) speed and the slow (40 mph) speed. This 6 minutes is 1/10 of an hour.
Distance = rate x time. The two distances are equal so:
We can figure out how long it would take at 60 mph and go from there.
40 (t + .1) = 60 t
40t + 4 = 60 t
4 = 20 t
4/20 = t = 1/5 hour
So it will take 1/5 hour at 60 mph which is 12 miles. (1/5 hour = 12 minutes)
It will take abit longer at 40 mph:
1/5 hour + 1/10 hour =
.2 hour + .1 hour = .3 hour at 40 mph which is also 12 miles. (.3 hour is 18 minutes).
If Ms. Brown was exactly on time it would take her 15 minutes to go the 12 miles.
12 miles in 15 min = 48 mph.